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Why is the probability of intersection of two independent sets $A$ and $B$, a multiplication of their respective probabilities i.e. Why is $$\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)?$$

this question is about the intuition behind the definition of independence of sets in a probability space

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    $\begingroup$ I disagree with the closing votes; this question is about the intuition behind the definition of independence of sets in a probability space, and as such, it's a valid question, if not strictly mathematical. Granted, OP should make it more clear that it's the intuition he's asking about. $\endgroup$
    – tomasz
    Oct 21, 2012 at 19:06
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    $\begingroup$ This question popped up on the first page two weeks after an answer got accepted. I am rather surprised to see that no explanation of the multiplicative property by enumeration processes in Cartesian products is even mentioned although this is clearly the basis of intuition about the independence property. Is the OP still interested? $\endgroup$
    – Did
    Nov 18, 2012 at 11:18
  • $\begingroup$ Maybe other people will be interested in the future. $\endgroup$ Mar 8, 2018 at 13:49

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The fact that the probability of the intersection of independent events $A$ and $B$ is the product of their probabilities is actually the definition of independent events.

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  • $\begingroup$ That still doesn't state the proof. I don't know if this is my lack of understanding how multiplication works, but ever since kindergarten multiplication has been known as A*B. where A is added B times to itself. Which clearly does not signify the relationship of intersection $\endgroup$ Oct 21, 2012 at 6:20
  • $\begingroup$ You're confused. We're talking about multiplying probabilities, which are numbers between $0$ and $1$. For example, $0.3 \times 0.6 = 0.18$. This is not "$0.3$ is added $0.6$ times to itself". $\endgroup$ Oct 21, 2012 at 16:25
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    $\begingroup$ I know it's not , but i mean to say is this common sense. Like if you had to find the union of two independent sets , its obvious, that it is addition with the removal of the number of intersecting terms. How did they define, they must had some knowledge in order to define this fact? $\endgroup$ Oct 21, 2012 at 19:02
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    $\begingroup$ Maybe this example will make it seem more "common-sense". Suppose the probability of a person being left-handed is $1/10$, and the probability of being blue-eyed is $1/3$. The intersection of the events "left-handed" and "blue-eyed" is "left-handed and blue-eyed". If the events are independent, the proportion of blue-eyed people among the left-handed people is the same as the proportion of blue-eyed people among the whole population, i.e. $1/3$. Thus $1/3$ of $1/10$, or $(1/3) \times (1/10) = 1/30$, of the population is left-handed and blue-eyed. $\endgroup$ Oct 21, 2012 at 19:37
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    $\begingroup$ thanks that makes sense! :D $\endgroup$ Oct 22, 2012 at 3:06
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If

  • half the slices of a pizza have anchovies ($P(A)=\frac12$), and
  • you take a third of the slices of pizza ($P(B)=\frac13$) independently of whether they have anchovies, then
  • the anchovy slices that you have are one-sixth of all the slices of pizza ($P(A\cap B) = \frac16$).

This is because if your taking of slices is truly independent of their having anchovies, then

  • you will take a third of the anchovy slices ($P(A\cap B) = \frac13 P(A)$) and a third of the non-anchovy slices;

  • equivalently, half the slices you have will have anchovies ($P(A\cap B) = \frac12 P(B)$) and half will not.

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  • $\begingroup$ Thanks, this was really helpful. $\endgroup$ Oct 22, 2012 at 3:07
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    $\begingroup$ The only thing that I don't get about this answer is why "if you're taking slices truly independent of their anchovies" then you'll take half 'non', and half 'with'. I don't follow the logic there because probability would imply that it's random, and that you may end up choosing only those 'with' just by chance. Would be cool if that logic were explained. $\endgroup$
    – user305437
    Oct 27, 2016 at 18:24
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    $\begingroup$ @Jill: You're right, the phrase "on average" should be sprinkled liberally throughout this answer. $\endgroup$
    – user856
    Oct 27, 2016 at 19:11
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Let A,B be two events. We say that A and B are independent of each other iff:

  • $\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)$

Now, note that A and B are independent of each other if and only if $\mathbb{P}(A|B) = \mathbb{P}(A)$. In other words, A and B are independent of each other if and only if the realization of one of the events does not affect the conditional probability of the other. Assume that we perform two random experiments independent of each other, meaning that the two experiments do not interact. That is, the experiments have no in influence on each other. Let A denote an event related to the first experiment, and let B denote an event related to the second experiment. We can see that in this situation the equation $\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)$ must hold in order for it to be independent. Thus, we have that A and B are independent if and only if it satisfies the above definition.

There are also many other cases where events related to a same experiment are independent, in the sense of the above definition. For example for a fair die, the events A = {1,2} and B = {2, 4, 6} are independent. Lets also say if two disjoint events intersection/union probability is equal to zero, in this case you will know they are dependent of one another because then if one event occurs the other doesn't hence dependency. There can also be more than two independent events at a time.

Here read this, hopefully it will make more sense http://en.wikipedia.org/wiki/Conditional_probability.

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  • $\begingroup$ Thankyou, I understand this , I just don't understand, why we use multiplication to show the probability of events that is similar to both the sets. Imagine if a four year old kid ask why is multiplication of P(A) and P(B) , where A and B are independent sets, considered finding the probability of their intersection. Is there a proof for this definition . How will you make the kid understand, why multiplication is used? $\endgroup$ Oct 21, 2012 at 19:05
  • $\begingroup$ I have edited my post with a bit more detail. Read the article that I provided, hopefully it will make more sense. $\endgroup$
    – diimension
    Oct 21, 2012 at 19:33
  • $\begingroup$ Also, I do not know of any proof of this particular definition. Sorry for that $\endgroup$
    – diimension
    Oct 21, 2012 at 19:41
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Perhaps, one way of looking at it is the fact that intersection means to add conditions. In that sense, for every individual satisfying the condition A you have to compute how many satisfy condition B, thus the multiplication issue. Note that satisfying two conditions is scarcer than satisfying just one, so the "weight" of individuals satisfying both conditions respect the whole population must be less than the weight of the ones satisfying just one condition. Then, given the fact that probabilities are less than or equal one, multiplication seems a reasonable way of describing the process.

I know it is not the same in a formal sense, but is like saying for every row, compute how many columns, then you get the total number of pieces.

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A proof is not necessary here, since this is a definition. If you want to understand this definition, consider an example.

Observe all the girls of a small town. There are 1000 girls: 100 of them has a dog and 250 of them are blonde.

So now what is the probability of "a girl is blonde and has a dog" ? We assume here that the fact a girl is blonde is irrelevant to know if she has a dog or not. In term of probability, the events A "a girl is blonde" and B "a girl has a dog" are assumed independent. So, the 250 blond girls are just a subset of the 1000 girls. Since each one of them has the same probability to have a dog than every other girl (blonde or not) we can think that around 25 blond girls has a dog.

In term of probability, $p(A \cap B) = 0.025 = 0.1 \times \frac{250}{1000}= p(A)p(B)$. Et voila !

Like diimension explains, we can also said that $p(B|A) = p(B)$.

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hold two transparent sheets (A and B) on top of each other with an overlapping area C .
when we say intersection of A and B (sheet A independent of sheet B) it means from every 'unique' part of sheet A ,overlapped by sheet B (CA) we can draw a 'unique' line to every part of sheet B ,over lapped by sheet A (CB). ...same goes when you say 5 * 6 you mean for every part of 5 there is all parts of 6

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in case of finding the intersection that both the events has occurred i.e. event A has occurred and then event B given A has already occurred will come into picture i.e. P(A ∩ B) = P(A) * P(B|A) but in case of independent sets probability of b given a has already occurred will result in P(B|A)=P(B) only, as B is independent of occurrence of A that's why P(A ∩ B) = P(A) * P(B) in case of independent events.

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The sample space of independent events to occur will be the Cartesian product of their individual sample space and expected event will also be the Cartesian product of their possible outcomes. Let m, n = cardinality of sample spaces and o1 , o2 be the cardinality of possible outcomes. Thus their probability of occurance will be (o1×o2)÷(m×n) and can be arranged as ( o1÷m) ×(o2÷n) i.e. p1×p2.

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You might have already understood the intuition behind it. But here is my view on it.

Conditional Probability: Probability of an event A given that event B happens. That is, Probability of A given B represented as $$P(A|B)= \frac{P(A \cap B)}{P(B)}$$

There lies the subtle idea in the definition of this. Given that event B happened, our sample space got reduced to "number of favourable outcomes for B". Now, we need to find the outcomes in which both A and B occurs given B occurs definitely i.e.,

$$P(A|B)= \frac{\text{Favourable outcomes for the occurrence of both $A$ and $B$}}{\text{Favourable outcomes for $B$}}$$ $$= \frac{\large \frac{\text{Favourable outcomes for the occurrence of both $A$ and $B$}}{\text{All possible outcomes}}}{\large \frac{\text{Favourable outcomes for $B$}}{\text{All possible outcomes}}}$$ $$= \frac{P(A \cap B)}{P(B)}$$

Now, if A and B are independent events, then there is no way B affects A. So, probability of occurrence of A does't change with respect to B, which means

$$P(A|B)=P(A)$$

Combining above two equations we have

$$\frac{P(A \cap B)}{P(B)} = P(A) \Rightarrow P(A \cap B) = P(A)\times P(B) $$

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before I give the proof---lemme me mention about conditional probability---.(..if you already know skip this para) its represented by P(E/F) it means-->Probability of event E given Event F has already occurred. Now if F has occurred the sample space of even E will 'reduce' or get limited to outcomes of F. eg: While tossing a die... 1,2,3,4,5,6 is the sample space. now if event of getting an even number has already occurred then find the probability of getting 2---- since the number on the die is surely an even no: the sample space reduces to 2,4,6 ...so P(E/F)=p(E[intersection)F)/P(F) ; {we take intersection....cuz we need only what is common in both the events ..because event F surely has occurred)

Now here is the proof.... for independent events E and F ....whether event E has occurred or not --it doesn't depend on event F So p(E/F)=p(e[intersn]F)/p(F)=p(E) ===>> p(E[intersc]F)=p(E)*p(F) for independant events

it doesn't simply come from the definition ...see there's the proof!!

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